Three concrete equivalences

Determine whether the following relations are equivalences and if they are, determine their equivalence classes.

Variant
\(X_1={\mathbb N}, xR_1y \Leftrightarrow p |(x-y)\) (residue classes modulo \(p\in {\mathbb N}, p\ge 2\))
Solution
Reflexivity: \(x-x=0\) and \(p|0\) for all \(p\).

Symmetry: if \(p |(x-y)\), then \(p |(y-x)\).

Transitivity: if \(p |(x-y)\) and \(p |(y-z)\), then \(p |(x-y)+(y-z)=(x-z)\).
Answer
This is an equivalence relation. The equivalence classes are \([a]_{R_1}=\{kp+a, k\in \mathbb N\}\) for \(a\in{0,…,p-1}\).
Variant
\(X_2={\mathbb Z}\setminus 0, xR_2y \Leftrightarrow x|y \wedge y|x\)
Solution
Reflexivity: \(x|x\) for all \(x\ne 0\).

Symmetry: if \(x|y \wedge y|x\), then \(y|x \wedge x|y\), because the logical operation "and" is commutative.

Transitivity: if \(x|y \wedge y|x\) and \(y|z \wedge z|y\), then combining the first and third terms yields \(x|z\), and the second and fourth yield \(z|x\).
Answer
This is an equivalence relation. The equivalence classes are \([a]_{R_2}=\{a,-a\}\) for \(a\in \mathbb N\).
Variant
\(X_3={\mathbb N}, xR_3y \Leftrightarrow \exists z\in {\mathbb N}: z|y \wedge z|x\).

What happens if we require \(z>1\)?
Solution
Because 1 divides all natural numbers, \(R_3\) has only one equivalence class, namely \(\mathbb N\). (All pairs belong to the relation, so all three equivalence properties are trivially satisfied.)

For \(z>1\) this is not an equivalence relation, e.g. \((2{,}6),(6{,}3)\in R_3\), but \((2{,}3)\notin R_3\).